Mathematics • Year 8 • Unit 4 • Lesson 9

Median and Mode, Mixed Challenge

Pull together everything from Lesson 9: finding median (odd/even n), identifying mode (none/one/bimodal), comparing measures of centre, and choosing the right one for the context.

Master · Mixed Challenge

1. Mixed problems, choose the right move

Each question uses a different idea from Lesson 9. Show your working. 3 marks each

1.1 Find the median, mode, and range of: 22, 18, 25, 22, 19, 28, 22, 24, 20.

1.2 Identify the type of mode for each (no mode / unimodal / bimodal):
(a) 1, 2, 3, 4, 5, 6.
(b) 2, 4, 4, 6, 8, 10.
(c) 1, 1, 2, 3, 3, 5, 5, 5.
(d) 3, 3, 5, 5, 7.

1.3 A data set of 6 values has mean 12, median 11.5, mode 8. (a) What might this tell you about the data's shape? (b) Invent a possible data set of 6 values matching these statistics (use whole numbers).

1.4 For each scenario, decide which measure of centre is best (mean / median / mode) and justify in one sentence:
(a) The "average" sale price in a suburb with several luxury mansions.
(b) Choosing the most popular T-shirt size to manufacture more of.
(c) Calculating a student's term grade from 5 consistent test scores.

1.5 Find the median of an EVEN-sized data set: 12, 8, 14, 6, 10, 18, 15, 9. Show your ordering.

1.6 A 6-value data set has mean 10 and median 9.5. A 7th value of 100 is added; 100 is greater than every original value. (a) What is the new mean? (b) Can you determine the exact new median? Explain what can be concluded about its direction of change and why the amount is unknown.

Stuck on 1.6? Old total = 10 × 6 = 60. New total = 60 + 100 = 160. New mean = 160 ÷ 7. Because 100 is greater than every original value, the new median is the original 4th ordered value. It cannot be below the old median, but without the original values its increase is not determined and may be substantial.

2. Find the mistake

Another student attempted this median problem. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks

Problem: Find the median of: 17, 23, 11, 19, 25, 15.

Line 1: The data has 6 values, so n = 6.

Line 2: For an even n, median = average of the 2 middle values.

Line 3: Middle values (in original order) are the 3rd and 4th: 11 and 19.

Line 4: Median = (11 + 19) ÷ 2 = 15.

Line 5: Answer: median = 15.

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong.

(c) Write out the corrected working and the correct median.

Stuck? You MUST order the data first before identifying the middle values.

3. Open-ended challenge, design a data set with target statistics

This question has many valid answers. 4 marks

3.1 Your job: invent a data set of EXACTLY 9 whole-number values that satisfies ALL of the following at once:
• mean = 20
• median = 18
• mode = 15 (single mode, appearing at least 3 times)
• range = 30

Write up your answer with:
(i) List your 9 values (ordered smallest to largest).
(ii) Verify the mean by showing the sum and division.
(iii) Verify the median by identifying the middle value.
(iv) Verify the mode by counting frequencies.
(v) Verify the range (max − min).

Stuck? Try starting with three 15s (for mode), put 18 in the 5th position (median), and arrange the rest so they sum to 180 and the range spans 30. One option: 5, 15, 15, 15, 18, 19, 22, 36, but that's 8 values, so adjust to fit 9 values summing to 180. Trial-and-error is allowed.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

1.1, Median, mode, range

Ordered: 18, 19, 20, 22, 22, 22, 24, 25, 28. n = 9. Median = 5th value = 22. Mode = 22 (3 times). Range = 28 − 18 = 10.

1.2, Type of mode

(a) No mode (each value appears once). (b) Unimodal (4 appears twice; one mode = 4). (c) Unimodal (5 appears 3 times, only one most-frequent value). (d) Bimodal (3 and 5 both appear twice).

1.3, Interpreting summary stats

(a) Mean (12) > median (11.5) > mode (8) suggests skewed right a few high values pull the mean up.
(b) Sample data set: 8, 8, 11, 12, 13, 20 (mean = (8+8+11+12+13+20)/6 = 72/6 = 12 ✓; median = (11+12)/2 = 11.5 ✓; mode = 8 ✓).

1.4, Choose the measure

(a) Median luxury mansions are outliers that inflate the mean.
(b) Mode the most popular size sells the best.
(c) Mean consistent scores with no outliers means the mean fairly represents the term average.

1.5, Even-sized median

Ordered: 6, 8, 9, 10, 12, 14, 15, 18. n = 8. Median = (4th + 5th) ÷ 2 = (10 + 12) ÷ 2 = 11.

1.6, Adding an outlier

(a) Old total = 60. New total = 160. New mean = 160 ÷ 7 ≈ 22.86 increased by ~13.
(b) Because 100 is greater than every original value, the new median is the original 4th ordered value. It cannot decrease; its exact change is not determined by the original mean and median, and it may be substantial.

2, Find the mistake

(a) The mistake is on Line 3.
(b) The student took the 3rd and 4th values from the data in its ORIGINAL order. The median requires the data to be ORDERED first; otherwise the "middle" values are arbitrary.
(c) Correct: order the data → 11, 15, 17, 19, 23, 25. 3rd and 4th values = 17 and 19. Median = (17 + 19) ÷ 2 = 18.

3, Open-ended (sample solution)

(i) Sample data set: 5, 12, 15, 15, 18, 21, 24, 35, 35... wait, that gives two modes. Try: 5, 15, 15, 15, 18, 19, 25, 32, 36. Check: range = 36 − 5 = 31 (not 30). Adjust to: 5, 15, 15, 15, 18, 20, 25, 32, 35. Range = 30 ✓.
(ii) Mean check: Σx = 5+15+15+15+18+20+25+32+35 = 180. Mean = 180 ÷ 9 = 20 ✓.
(iii) Median check: n = 9, position 5 = 18 ✓.
(iv) Mode check: 15 appears 3 times, most frequent → mode = 15 ✓.
(v) Range check: 35 − 5 = 30 ✓.

Marking: 1 mark for any valid 9-value data set; 1 mark for correct mean verification; 1 mark for correct median + mode verification; 1 mark for correct range verification. Note: students can use trial-and-error to find any valid set.